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I have recently learned of the Monty Hall problem, where given three doors, you choose one with a 1/3 probability of choosing the correct door, and then switching to the other door changing the probability of choosing the correct door to 1/2. I am confused because wouldn't staying at the same door have an equal probability at 1/2 because by staying at that door you are effectively choosing that door in the smaller sample space?

Thanks for the help!

Jam
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John Kent
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    Imagine thet you had 100 doors. I open 98 of them. I don't open them arbitrarily. I know where the prize is. Would you switch now? – John Douma Jul 11 '17 at 20:17
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    Consider this: The game show host always opens an incorrect door after the player's initial choice, leaving the door the player initially chose and one other door (one of which is the correct door). The only way switching the choice of door could lead to the player not choosing the correct door is if they had already chosen the correct door, which only happens with probability $1/3$. – Michael L. Jul 11 '17 at 20:18

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Probability relies heavily on the knowledge of the observer. The question might be, "What's the probability a randomly chosen man is taller than 6 feet?$ If I know nothing more about him, the answer is one thing. If I know he's a basketball star, then the answer will be different.

There are two observers lurking around the Monty Hall problem. The player and the hypothetical alien who walks in later after the first door is chosen and after Monty has revealed a goat behind one of the doors. To the alien, with his limited knowledge, the probability is 1/2 that the car is behind either door. But to the player, who knows a bit more information, the probability is different. He knows that he had a 1/3 chance with his first choice. So he knows that there's a 2/3 chance the car is behind one of the other two doors. If it is, then, by the rules of the game, he'll win the car by switching.

B. Goddard
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  • This makes sense, so probability is not a property of the state of a system, but rather the knowledge an observer has of the state? – John Kent Jul 11 '17 at 20:32
  • @JohnKent It doesn't really depend on there being an observer, exactly. It is about the rules of the game. #1: 2/3 of the time your first guess was wrong. #2: Monty will never reveal a car. You need both of these. If Monty could open a door to reveal a car but forbid you to choose that newly opened door, then your advantage vanishes. – Ian Jul 11 '17 at 20:51
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The Monty Hall problem is only confusing if you insist on modeling it as the probability that the prize is behind door number $2$ given that it is not behind door number $1$. This doesn't apply here because Monty doesn't open doors arbitrarily. He knows where the prize is. If you initially guess right, he opens any of the other two doors. If you don't, he opens the other door that doesn't have the prize.

That said, what you really want is to compare two conditional probablilities:

  1. $P($Win|Stay$)$
  2. $P($Win|Switch.$)$

$P($Win|Stay$)=\frac{1}{3}$ because you only win if you were originally right which happens $\frac{1}{3}$ of the time.

$P($Win|Switch.$)=\frac{2}{3}$ because you win if you were originally wrong which happens $\frac{2}{3}$ of the time.

Therefore, you maximize your chance of winning by switching.

John Douma
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You have two strategies: Switch doors or not to switch. The strategy can be chosen beforehand as nothing in the game dynamics will add any valuable information to you. Monty will always open one of the others doors and it will be empty. Thus, there are two possible situation:

Under switching strategy. Using the law of total probability you have that
$$ P(\text{Win}) = P(\text{Win}|\text{Right})P(\text{Right}) + P(\text{Win}|\text{Wrong})P(\text{Wrong}) = \frac{1}{3}\times0 + \frac{2}{3}\times1=\frac{2}{3}. $$

While under non-switching strategy. Using the law of total probability you have that
$$ P(\text{Win}) = P(\text{Win}|\text{Right})P(\text{Right}) + P(\text{Win}|\text{Wrong})P(\text{Wrong}) = \frac{1}{3}\times 1 + \frac{2}{3}\times0=\frac{1}{3}. $$

As such, by switching you are maximizing your probability to win.

V. Vancak
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